Question

Let the centre of a circle, passing through the point $(0, 0)$, $(1, 0)$ and touching the circle $x^2 + y^2 = 9$, be $(h, k)$. Then for all possible values of the coordinates of the centre $(h, k)$, $4(h^2 + k^2)$ is equal to __________.

Answer 1

Step 1: General equation of the circle The equation of the circle is:

$(x - h)^2 + (y - k)^2 = r^2.$

Since the circle passes through the points $(0, 0)$ and $(1, 0)$, substitute these points into the circle’s equation.

For $(0, 0)$:

$h^2 + k^2 = r^2.$

For $(1, 0)$:

$(1 - h)^2 + k^2 = r^2.$

Expanding and simplifying:

$1 - 2h + h^2 + k^2 = h^2 + k^2.$

Substitute $r^2 = h^2 + k^2$ into the equation:

$1 - 2h + h^2 + k^2 = h^2 + k^2.$

Cancel $h^2 + k^2$:

$1 - 2h = 0 \implies h = \frac{1}{2}.$

Step 2: Circle touches $x^2 + y^2 = 9$ The given circle $x^2 + y^2 = 9$ has a radius $R = 3$ and is centered at $(0, 0)$. For the circle to touch $x^2 + y^2 = 9$, the distance between their centers must be equal to the difference of their radii:

$\sqrt{h^2 + k^2} = R - r = 3 - \sqrt{h^2 + k^2}.$

Let $d = \sqrt{h^2 + k^2}$:

$d = 3 - d \implies 2d = 3 \implies d = \frac{3}{2}.$

Thus:

$h^2 + k^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}.$

Step 3: Compute $4(h^2 + k^2)$ Multiply $h^2 + k^2$ by 4:

$4(h^2 + k^2) = 4 \cdot \frac{9}{4} = 9.$

Final Answer: 9.